A projectile is shot from the ground at an angle of #(5 pi)/12 # and a speed of #5 m/s#. Factoring in both horizontal and vertical movement, what will the projectile's distance from the starting point be when it reaches its maximum height?

Question

1390 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-07-04T18:48:18

The distance is #=1.349m#

The equation of the trajectory of the projectile in the vertical plane, #O# being the origin is

#y=xtantheta-((g)/(2u^2cos^2theta))x^2#

The initial velocity is #u=5ms^-1#

The angle of projection is #theta=5/12pi#

The acceleration due to gravity is #g=9.8ms^-2#

Therefore,

#y=xtan(5/12pi)-((9.8)/(2*5^2*cos^2(5/12pi)))x^2#

#y=3.732x-2.926x^2#

The highest point is reached when #dy/dx=0#

#dy/dx=3.732-5.852x#

#3.732-5.852x=0#

#=>#, #x=3.732/5.852=0.638#

And

#y=3.732*0.638-2.926*0.638^2=1.189#

The highest point is #=(0.638,1.189)#

The distance from the stat¡rting point is

#d=sqrt(0.638^2+1.189^2)=1.349m#

graph{(3.732x-2.926x^2) [-0.988, 2.43, -0.128, 1.581]}